Thom class

Let h* be a generalized cohomology theory (for example, let h*=H*, singular cohomology with integer coefficients). Let ξ→X be a vector bundle of dimensiond over a topological spaceX. Assume for convenience that ξ has a Riemannian metric, so that we may speak of its associated sphere and disk bundles, S⁢(ξ) and D⁢(ξ) respectively.

Let x∈X, and consider the fibers S⁢(ξx) and D⁢(ξx). Since D⁢(ξx)/S⁢(ξx) is homeomorphic to the d-sphere, the Eilenberg-Steenrod axioms for h* imply that h*+d⁢(D⁢(ξx),S⁢(ξx)) is isomorphic to the coefficient group h*⁢(pt) of h*. In fact, h*⁢(D⁢(ξx),S⁢(ξx)) is a free module of rank one over the ring h*⁢(pt).

Definition 2

If a Thom class for ξ exists, ξ is said to be orientable with respect to the cohomology theory h*.

Remark 1

Notice that we may consider τ as an element of the reducedh*-cohomology grouph~*⁢(Xξ), where Xξ is the Thom space D⁢(ξ)/S⁢(ξ) of ξ. As is the case in the definition of the Thom space, the Thom class may be defined without reference to associated disk and sphere bundles, and hence to a Riemannian metric on ξ. For example, the pair (ξ,ξ-X) (where X is included in ξ as the zero section) is homotopy equivalent to (D⁢(ξ),S⁢(ξ)).

Remark 2

If h* is singular cohomology with integer coefficients, then ξ has a Thom class if and only if it is an orientable vector bundle in the ordinary sense, and the choices of Thom class are in one-to-one correspondence with the orientations.